Amy marks two points A and B that are 4 inches apart. She draws one circle that has segment AB as a diameter. She then draws a larger circle, which overlaps the first circle, such that the arc from A to B along its circumference is a quarter-circle. What is the total area covered by the two circles? Express your answer in terms of m.
The radius of the smaller circle is half the length of AB, so it is 2 inches.
The circumference of the smaller circle is 2π times the radius, so it is 4π inches.
The larger circle has a quarter-circle arc of AB, so its total circumference is four times longer than the circumference of the smaller circle, so it is 16π inches.
Divide the circumference by 2π to find the diameter is 8 inches.
So, the radius of the larger circle is half the diameter, or 4 inches.
The area of a circle is πr², so the area of the smaller circle is π(2²) = 4π square inches.
The area of the larger circle is π(4²) = 16π square inches.
The total area covered by the two circles is 4π + 16π = <<4+16=20>>20π square inches. Answer: \boxed{20\pi}.
To find the total area covered by the two circles, we need to calculate the areas of the individual circles and then add them together.
Let's first find the area of the smaller circle, which has segment AB as a diameter. The radius of this circle is half of the diameter, so the radius is 4 inches / 2 = 2 inches. The area of a circle is given by the formula A = πr^2, where A is the area and r is the radius. Plugging in the radius, we get A1 = π(2 inches)^2 = 4π square inches.
Next, let's find the area of the larger circle. Since the arc from A to B along its circumference is a quarter-circle, we can consider the diameter to be the diameter of the smaller circle plus the length of segment AB. The length of segment AB is given as 4 inches, so the diameter of the larger circle is 4 inches + 4 inches = 8 inches. The radius of the larger circle is half of the diameter, so the radius is 8 inches / 2 = 4 inches. The area of this circle is A2 = π(4 inches)^2 = 16π square inches.
Finally, we can calculate the total area covered by the two circles by adding the areas together. A_total = A1 + A2 = 4π square inches + 16π square inches = 20π square inches.
Therefore, the total area covered by the two circles is 20π square inches.
To find the total area covered by the two circles, let's break down the problem step by step.
1. Start by finding the radius of the smaller circle. Since the distance between points A and B is 4 inches and AB is the diameter of the smaller circle, we can deduce that the radius (r) is half of the diameter, which is 2 inches.
2. Next, find the area of the smaller circle using the formula for the circle's area: A = πr^2. In this case, the area (A1) of the smaller circle would be A1 = π(2 inches)^2.
3. Calculate the area of the quarter-circle. The quarter-circle is formed by an arc connecting points A and B. This arc is a quarter of the circumference of the larger circle. Since we know the diameter of the larger circle is AB, which is 4 inches, the radius (R) can be calculated by dividing the diameter by 2: R = 4 inches / 2 = 2 inches.
4. The circumference of the larger circle (C2) is calculated using the formula C = 2πR. In this case, C2 = 2π(2 inches).
5. As the arc from A to B is a quarter of the circumference, the length of the arc is a quarter of C2. Hence, the length of the arc (l) is given by l = 1/4 * C2.
6. The area of the quarter-circle (A2) can be calculated using the formula for the area of a sector: A2 = (θ/360) * πR^2, where θ is the central angle of the sector. Since our sector is a quarter of a circle, the central angle is 90 degrees.
7. Combining the above steps, we have A2 = (90/360) * π(2 inches)^2. Simplifying this expression will give us the value of A2.
8. Finally, to find the total area covered by the two circles, we add the areas of the smaller circle (A1) and the quarter-circle (A2): Total Area = A1 + A2.
By following the above steps and plugging in the appropriate values for the radii and angles, you can calculate the total area covered by the two circles in terms of m.